Molarity is moles of solute per litre of solvent. We're given mass of solute, but can convert to moles using the molar mass of NaOH (40 g/mol). There are 10 g / 40 g/mol = 0.25 mol of NaOH in the solution.
M = 0.25 mol / 2.0 L = 0.125M
The second answer (0.13M) is the correct one.
The answer for the following problem is mentioned below.
- <u><em>Therefore the final temperature of the gas is 740 K</em></u>
Explanation:
Given:
Initial pressure of the gas (
) = 1.8 atm
Final pressure of the gas (
) = 4 atm
Initial temperature of the gas (
) = 60°C = 60 + 273 = 333 K
To solve:
Final temperature of the gas (
)
We know;
From the ideal gas equation;
we know;
P × V = n × R × T
So;
we can tell from the above equation;
<u> P ∝ T</u>
(i.e.)
<em> </em>
<em> = constant</em>
= 
Where;
= initial pressure of a gas
= final pressure of a gas
= initial temperature of a gas
= final temperature of a gas
= 
=
= 740 K
<u><em>Therefore the final temperature of the gas is 740 K</em></u>
A word equation is a chemical reaction described using words.
A common example is the act of photosynthesis - the process plants use to make glucose (sugar) to use as 'food'.
Plants convert water and carbon dioxide into oxygen and glucose.
A word equation to express this is:
Water + Carbon Dioxide → Glucose + Oxygen
The other type of equation is a symbol equation - this uses the symbols of the elements instead of the common names:
H₂O + CO₂ → C₆H₁₂O₆ + O₂
There is also a balanced version:
6H₂O + 6CO₂ → C₆H₁₂O₆ + 6O₂
<em>If you want information on the balanced symbol equations, feel free to PM me.</em>
Answer:
The Answer should be Endothermic reaction
Explanation:
Reactions in which energy is absorbed are called Endothermic reactions.
Answer:
54 days
Explanation:
We have to use the formula;
0.693/t1/2 =2.303/t log Ao/A
Where;
t1/2= half-life of phosphorus-32= 14.3 days
t= time taken for the activity to fall to 7.34% of its original value
Ao=initial activity of phosphorus-32
A= activity of phosphorus-32 after a time t
Note that;
A=0.0734Ao (the activity of the sample decreased to 7.34% of the activity of the original sample)
Substituting values;
0.693/14.3 = 2.303/t log Ao/0.0734Ao
0.693/14.3 = 2.303/t log 1/0.0734
0.693/14.3 = 2.6/t
0.048=2.6/t
t= 2.6/0.048
t= 54 days
